Python递归地替换嵌套字典的键中的字符？

问题描述

我正在尝试创建一个通用函数，用于替换嵌套字典的键中的点。我有一个非泛型功能，深入3级，但必须有一种方法来做这个泛型。任何帮助是赞赏！我的代码到目前为止：

I'm trying to create a generic function that replaces dots in keys of a nested dictionary. I have a non-generic function that goes 3 levels deep, but there must be a way to do this generic. Any help is appreciated! My code so far:

output = {'key1': {'key2': 'value2', 'key3': {'key4 with a .': 'value4', 'key5 with a .': 'value5'}}} 

def print_dict(d):
    new = {}
    for key,value in d.items():
        new[key.replace(".", "-")] = {}
        if isinstance(value, dict):
            for key2, value2 in value.items():
                new[key][key2] = {}
                if isinstance(value2, dict):
                    for key3, value3 in value2.items():
                        new[key][key2][key3.replace(".", "-")] = value3
                else:
                    new[key][key2.replace(".", "-")] = value2
        else:
            new[key] = value
    return new

print print_dict(output)

更新：为了回答自己的问题，我使用json object_hooks做了一个解决方案：

import json

def remove_dots(obj):
    for key in obj.keys():
        new_key = key.replace(".","-")
        if new_key != key:
            obj[new_key] = obj[key]
            del obj[key]
    return obj

output = {'key1': {'key2': 'value2', 'key3': {'key4 with a .': 'value4', 'key5 with a .': 'value5'}}}
new_json = json.loads(json.dumps(output), object_hook=remove_dots) 

print new_json

推荐答案

是的，存在更好的方法：

Yes, there exists better way:

def print_dict(d):
    new = {}
    for k, v in d.iteritems():
        if isinstance(v, dict):
            v = print_dict(v)
        new[k.replace('.', '-')] = v
    return new

（编辑：它的递归，更多在）。

( It's recursion, more on Wikipedia.)

这篇关于Python递归地替换嵌套字典的键中的字符？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！